can you break this down for me for the answer A balloon used for atmospheric research has a volume of 1.0 x 106 L. Assume that the balloon is filled with helium gas at STP (standard temperature and pressure) and then allowed to ascend to an altitude of 10 km, where the pressure of the atmosphere is 243 mm Hg and the temperature is -33°C. What will the volume of the balloon be under these atmospheric conditions? (Hint: Three variables have been given so what equation will you use?) According to the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature. Assuming that the amount of helium gas in the balloon remains constant, we can use this equation to solve for the final volume of the balloon at its new altitude. First, we need to convert the given pressure and temperature to SI units. 243 mm Hg is equivalent to 32,400 Pa (1 atm = 101325 Pa), and -33°C is equivalent to 240 K (0°C = 273.15 K). Next, we can rearrange the Ideal Gas Law equation to solve for V2: V2 = (nRT2) / P2 where T2 and P2 are the temperature and pressure at the new altitude. Since the number of moles of gas remains constant, we can use the ratio of temperatures and pressures to simplify this equation: V2 = V1 * (T2 / T1) * (P1 / P2) where V1 is the initial volume of the balloon at STP. Substituting in our values, we get: V2 = (1.0 x 10^6 L) * (240 K / 273 K) * (101325 Pa / 32400 Pa) V2 ≈ 3.06 x 10^6 L

Question

can you break this down for me for the answer A balloon used for atmospheric research has a volume of 1.0 x 106 L. Assume that the balloon is filled with helium gas at STP (standard temperature and pressure) and then allowed to ascend to an altitude of 10 km, where the pressure of the atmosphere is 243 mm Hg and the temperature is -33°C. What will the volume of the balloon be under these atmospheric conditions? (Hint: Three variables have been given so what equation will you use?) According to the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature. Assuming that the amount of helium gas in the balloon remains constant, we can use this equation to solve for the final volume of the balloon at its new altitude.  First, we need to convert the given pressure and temperature to SI units. 243 mm Hg is equivalent to 32,400 Pa (1 atm = 101325 Pa), and -33°C is equivalent to 240 K (0°C = 273.15 K).  Next, we can rearrange the Ideal Gas Law equation to solve for V2:  V2 = (nRT2) / P2  where T2 and P2 are the temperature and pressure at the new altitude.  Since the number of moles of gas remains constant, we can use the ratio of temperatures and pressures to simplify this equation:  V2 = V1 * (T2 / T1) * (P1 / P2)  where V1 is the initial volume of the balloon at STP.  Substituting in our values, we get:  V2 = (1.0 x 10^6 L) * (240 K / 273 K) * (101325 Pa / 32400 Pa)  V2 ≈ 3.06 x 10^6 L

Bot GPT 3.5 · Answer

To find the final volume of the balloon at 10 km altitude, we first need to convert the given pressure and temperature to SI units: 243 mm Hg is equivalent to 32,400 Pa and -33°C is equivalent to 240 K. Next, we can rearrange the Ideal Gas Law equation to solve for V2 (final volume): V2 = V1 * (T2 / T1) * (P1 / P2), where T2 and P2 are the temperature and pressure at the new altitude. Since the number of moles of gas remains constant, we can use the ratio of temperatures and pressures to simplify this equation. Substitute in the given values: V2 = (1.0 x 10^6 L) * (240 K / 273 K) * (101325 Pa / 32400 Pa) V2 ≈ 3.06 x 10^6 L Therefore, the final volume of the balloon at 10 km altitude will be approximately 3.06 x 10^6 L.